Let `I=int(dx)/(1+3sin^(2)x)=(1)/(2)tan^(-1)(2f(x))+C` (where, C is the constant of integration). If `f((pi)/(4))=1`, then the fundamental period of `y=f(x)` is

Options

A. `(pi)/(4)`
B. `pi`
C. `2pi`
D. `(pi)/(6)`

Answer

To solve the given problem step by step, we start with the integral provided:

### Step 1: Set up the integral
We have:
\[ I = \int \frac{dx}{1 + 3\sin^2 x} \]

### Step 2: Simplify the integral
To simplify the integral, we can multiply the numerator and denominator by \(\cos^2 x\):
\[ I = \int \frac{\cos^2 x \, dx}{\cos^2 x + 3\sin^2 x \cos^2 x} = \int \frac{\cos^2 x \, dx}{\cos^2 x + 3(1 - \cos^2 x)} = \int \frac{\cos^2 x \, dx}{4\cos^2 x - 3} \]

### Step 3: Use a substitution
Let \(t = \tan x\), then \(dt = \sec^2 x \, dx\) or \(dx = \frac{dt}{1 + t^2}\). We also know that \(\sin^2 x = \frac{t^2}{1+t^2}\) and \(\cos^2 x = \frac{1}{1+t^2}\):
\[ I = \int \frac{\frac{1}{1+t^2} \cdot \frac{dt}{1+t^2}}{1 + 3\frac{t^2}{1+t^2}} = \int \frac{dt}{(1+t^2)(1 + 3t^2)} \]

### Step 4: Partial fraction decomposition
We can decompose:
\[ \frac{1}{(1+t^2)(1+3t^2)} = \frac{A}{1+t^2} + \frac{B}{1+3t^2} \]
Multiplying through by the denominator and solving for \(A\) and \(B\) gives:
\[ 1 = A(1 + 3t^2) + B(1 + t^2) \]
Setting \(t = 0\) gives \(A + B = 1\). Setting \(t^2 = 1\) gives \(A + 3A = 1\) or \(4A = 1\) so \(A = \frac{1}{4}\) and \(B = \frac{3}{4}\).

### Step 5: Integrate each term
Now we can integrate:
\[ I = \frac{1}{4} \int \frac{dt}{1+t^2} + \frac{3}{4} \int \frac{dt}{1+3t^2} \]
Using the integral formula \(\int \frac{dt}{1+t^2} = \tan^{-1}(t)\) and \(\int \frac{dt}{1+3t^2} = \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t)\):
\[ I = \frac{1}{4} \tan^{-1}(t) + \frac{3}{4\sqrt{3}} \tan^{-1}(\sqrt{3}t) + C \]

### Step 6: Substitute back for \(t\)
Substituting back \(t = \tan x\):
\[ I = \frac{1}{4} \tan^{-1}(\tan x) + \frac{3}{4\sqrt{3}} \tan^{-1}(\sqrt{3}\tan x) + C \]
This simplifies to:
\[ I = \frac{1}{4} x + \frac{3}{4\sqrt{3}} \tan^{-1}(\sqrt{3}\tan x) + C \]

### Step 7: Relate to \(f(x)\)
Given that:
\[ I = \frac{1}{2} \tan^{-1}(2f(x)) + C \]
We can equate:
\[ \frac{1}{4} x + \frac{3}{4\sqrt{3}} \tan^{-1}(\sqrt{3}\tan x) = \frac{1}{2} \tan^{-1}(2f(x)) + C \]

### Step 8: Find \(f(x)\)
From the equation, we can find \(f(x)\):
\[ f(x) = \frac{1}{2} \tan\left(\frac{x}{2}\right) \]

### Step 9: Determine the fundamental period
The function \(f(x) = \tan\left(\frac{x}{2}\right)\) has a fundamental period of:
\[ \text{Period} = 2\pi \div \frac{1}{2} = 4\pi \]

### Final Answer
Thus, the fundamental period of \(y = f(x)\) is:
\[ \boxed{4\pi} \]